Question

Prove\:\frac{\cos \left(2x\right)-\cos \left(4x\right)}{\sin \left(2x\right)+\sin \left(4x\right)}=\tan \left(x\right)

Answer 1

Prove
\[\frac{ \cos2x-\cos4x }{ \sin2x+\sin4x }=tanx\]

Answer 2

mhmm, I would say to do a substitution

2x = y

\[\frac{ \cos y-\cos2y }{ \sin y+\sin2y }=tan\frac{y}{2}\]

Using sin(2x) = 2sin(x)cos(x)
\[\frac{ \cos y-\cos2y }{ \sin y+2\sin y \cos y}=tan\frac{y}{2}\]

\[\frac{ \cos y-\cos2y }{ \sin y(2\cos y + 1)}=tan\frac{y}{2}\]

Using cos(2x) = 2cos^2(x) – 1
\[\frac{ \cos y-2\cos^2(y) – 1 }{ \sin y(2\cos y + 1)}=tan\frac{y}{2}\]

\[\frac{ -\cos y(2\cos y +1) }{ \sin y(2\cos y + 1)}=tan\frac{y}{2}\]

\[\frac{ -\cos y }{ \sin y}=tan\frac{y}{2}\]

I'm sure you can handle it from there. And don't forget we did 2x = y

Answer 3

Wait, I don't know what to do from there x'D

I would say to use

\(\Large tan\frac{y}{2} = \frac{sin(y)}{1+cos(y)}\)

and then cross multiply to prove it?

I dunno xD

Answer 4

@TheSmartOne you must have made a mistake http://www.wolframalpha.com/input/?i=%5Cfrac%7B+-%5Ccos+y+%7D%7B+%5Csin+y%7D%3Dtan%5Cfrac%7By%7D%7B2%7D

Answer 5

here \[\large \frac{ \cos y-(2\cos^2(y) – 1) }{ \sin y(2\cos y + 1)}=\tan\frac{y}{2}\]

Answer 6

ohhh! ooopsiees

Using cos(2x) = 2cos^2(x) – 1
\[\frac{ \cos y-(2\cos^2(y) – 1 )}{ \sin y(2\cos y + 1)}=tan\frac{y}{2}\]

\[\frac{ \cos y(1 -(2\cos^2(y) -1)) }{ \sin y(2\cos y + 1)}=tan\frac{y}{2}\]


\[\frac{ \cos y(-2\cos y +2) }{ \sin y(2\cos y + 1)}=tan\frac{y}{2}\]


mhmm

Answer 7

I was lazy/tired to try factoring...
http://www.wolframalpha.com/input/?i=cos+y+-+2cos%5E2+y+%2B+1+alternate+forms

\[\large \frac{ \cos y-(2\cos^2(y) – 1 )}{ \sin y(2\cos y + 1)}=\tan\frac{y}{2}\] \[\large \frac{ -(\cos y - 1)(2 \cos y +1)}{ \sin y(2\cos y + 1)}=\tan\frac{y}{2}\] \[\large \frac{ 1- \cos y}{ \sin y}=\tan\frac{y}{2}\]and you're done
http://home.windstream.net/okrebs/Ch10-17.gif

Answer 8

just to be safe\[\large \frac{ 1- \cos (2x)}{ \sin (2x)}=\tan\frac{(2x)}{2}\]

Answer 9

well done xD

Answer 10

now to transfer some of the medals I got to you ;p

Answer 11

You too! Though... i have no idea what you were doing in the post above mine... x'd

Answer 12

I was too sleepy to figure out what to do xD

Answer 13

\[\cos C-\cos D=2\sin \frac{ C+D }{ 2 }\sin \frac{ D-C }{ 2 }\]
\[\sin C+\sin D=2\sin \frac{ C+D }{ 2 }\cos \frac{ C-D }{ 2 }\]